LG Game 2 Question 11 Explanation, by LSATHacks
This question adds a new rule. As with question 10, I drew a new diagram using the additional rule:
Well, that didn’t seem to do much. But are any other rules affected? Rule 3 is the only rule not on this diagram:
On this question, G is not after H. So that means that M can’t be before P. Thus, P is before M. We can redraw the diagram above to show that:
Honestly, it’s an awkward diagram. On my own page I had arced lines, and it was a bit clearer. It’s also potentially confusing: Visually, H is to the right of J. But, there is no line connecting them! So, J could be after H.
There might be a better way to draw that, but I can’t think of one. Do enough linear games and you’ll learn to read this – only connecting lines matter, not position.
I should also note that I didn’t draw this diagram in two steps, as I’ve done in this explanation. I just drew the diagram all at once. If you can think of the deduction with rule 3 before drawing, that’s the best way. Otherwise, you should go step by step.
Now go through the answers and use the diagram to eliminate them:
- A could be false if you put both P and M before LG.
- B doesn’t have to be true: J could go last. There’s no line connecting H and J.
- C is CORRECT. Only P and M can go before L, which would make L third at latest.
- D and E are wrong for the same reason: either LG or P – M could go first. So neither answer has to be true.
I think there is a more straightforward answer to this question. If you take the standard diagram and add “G” immediately after L, there are two variables that follow: J and H. In which case, you can deduct that the latest L can go is third, because of the G – J / H train.
Why can only P and M go before L?
Because we’re told that LG are next to each other in that order, this means that H and J comes after them (since L needs to come before them both). Because M and P need to go before H and J, they can go anywhere before, even before L.